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Homework Help: Multiple delta solutions in limit proof

  1. Mar 11, 2010 #1
    Just wondering why there can be more than one delta solution for epsilon/delta proofs of a limit. I mean, shouldn't there just be one delta that can be used at any point on the function? Why is it that there is more than one and you choose the minimum delta..? I saw a geometric representation of why you would choose the minimum (if part of the function was near an asymptote) but can't you find one so that even at the asymptote it will be within epsilon? I hope I am making sense.
  2. jcsd
  3. Mar 11, 2010 #2


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    First of all, different proofs may use different delta's, but at any point there is a maximum delta, say [itex]\delta_0[/itex], such that
    [itex]|x - a| < \delta \implies |f(x) - L| < \epsilon[/itex] if and only if [itex]\delta < \delta_0[/itex]. But since the definition only states that there exists some delta, it is not necessary to find this bound (i.e. if you find a smaller delta than absolutely necessary, you can still prove the limit).

    Or were you asking: why is it possible that the delta we need depends on the point in which we are taking the limit? In that case, you might want to read up about different forms of continuity. For example, we say that a function is continuous at a, if
    [tex]\forall \epsilon, \forall a, \exists \delta = \delta(a): |x - a| < \delta \implies |f(x) - f(a)| < \epsilon[/tex]
    and then a function is continuous if it is continuous at all a (that is for all a and all epsilon we can find a delta which works for that epsilon and that a).
    We can also require that
    [tex]\forall \epsilon, \exists \delta, \forall a, |x - a| < \delta \implies |f(x) - f(a)| < \epsilon[/tex]
    That is, for all epsilon we can find a single delta which works for that epsilon, but for all a. Obviously, this is a stronger statement. For example, even if you can find a delta which depends on a, it may be possible that the minimum (or infimum) over all these deltas is zero. In that case the function is continuous, but you still cannot find a delta which works for all points at once.
    If such a delta does exist, we say that the function is uniformly continuous.

    About the question about the asymptote, it is not quite clear to me what you mean here. Do you mean: why can't we find a delta around, for example, x = 0 for f(x) = 1/x?
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